A218085 Let S_5(x) denote the difference in counts of multiples of 5 in the interval [0,x), those with even digit sums in base 4 in one set, those with odd digit sums in base 4 in the other. Then a(n) = (-1)^s_4(n) *(S_5(n) -10*S_5(floor(n/16)) +5*S_5(floor(n/256))), where s_4(n) = A053737(n).

0, -1, 1, -1, -1, 1, -2, 2, 2, -2, 2, -3, -3, 3, -3, 3, 6, -6, 6, -6, -6, 5, -5, 5, 5, -5, 4, -4, -4, 4, -4, 3, -3, 3, -3, 3, 4, -4, 4, -4, -4, 3, -3, 3, 3, -3, 2, -2, 2, -2, 2, -3, -3, 3, -3, 3, 4, -4, 4, -4, -4, 3, -3, 3, 3, -3, 2, -2, -2, 2, -2, 1, 1, -1, 1, -1, 0, 0, 0, 0, 0

Offset: 0

Author

Vladimir Shevelev and Peter J. C. Moses, Oct 20 2012

Keywords

• sign
• base
• easy

Comments

The sequence S_5(n) starts 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, ... for n >= 0. Apart from the initial 0, these are blocks of 5 repetitions of 1, 2, 3, 4, 5, 6, 7, 6, 7, 8, 7, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...

Theorem. The sequence is periodic with period 2560.

The theorem allows us to write a recursion for S_5(n), considering n modulo 2560: S_5(n) = 10*S_5(floor(n/16)) - 5*S_5(floor(n/256)) + (-1)^s_4(n)*a(n).

Examples

			a(n)=-9 for n=2411, 2412, 2414, 2491, 2492, 2494 (mod 2560);
a(n)=9 for n=2413, 2415, 2493, 2495 (mod 2560).
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 0..2559
• Vladimir Shevelev and Peter J. C. Moses, A family of digit functions with large periods, arXiv:1209.5705 [math.NT], 2012.

Crossrefs

Cf. A214458, A217971, A053737.

Programs

Maple

S := proc(n,j,x)
    a := 0 ;
    for r from j to x-1 by n do
        add(d,d=convert(r,base,n-1)) ;
        a := a+(-1)^% ;
    end do:
    a ;
end proc:
A218085 := proc(n)
    S(5,0,n)-10*S(5,0,floor(n/16))+5*S(5,0,floor(n/256)) ;
    %*(-1)^A053737(n) ;
end proc:
seq(A218085(n),n=0..80) ; # R. J. Mathar, Oct 31 2012

Formula

-9 <= a(n) <= 9, all 19 values are actually achieved.